## A building site and electrical equipment

**A building site is to have the following electrical equipment available for use:**

- Tower crane, electric motors totalling
**75 kW at 415 V**; - Sump pump,
**5 kW at 240 V**; - 60 tungsten lamps of
**100 W each at 240 V**; - 12 flood lamps of
**400 W each at 240 V**; - 20 hand tools of
**400 W at 110 V**.

**To calculate //**

- Find the
**total kilovolt-amperes**to be supplied to the site if the power factor of all rotary equipment is**0.8**. - Find the
**electrical current rating**for the incoming supply cable to the site. - Estimate
**the cost of electricity consumed**on the site during a 12-month contract.

### Ok, let’s do it…

**For rotating machinery, formulae for power in Volt Amperes [VA] goes like this:**

**For single-phase current:**

**For three-phase current:**

**The results of the power calculations are given in Table 1 below. The answers required are as follows.**

1. The total input power kilovolt-amperes required for site is **120.8 kVA**

2. The incoming supply cable capacity at 415 volt, three-phase, 50 Hz required is:

**This is not the same as a total of the currents**calculated in Table 1 as these larger numbers only appear at their reduced voltages in the relevant sub-circuits.

3. Assume that the crane, pump and tools are running for **25% of an 8-h working day**, 5 days per week for 48 weeks, 20 of the tungsten lamps are for security lighting 16 h every night, and the remaining 40 tungsten lamps and the flood lamps are used for 3 h per day, 5 days per week for the winter period of 20 weeks.

**The crane, pump and tools are working for:**

**The security lamps are working for:**

**The other lamps are working for:**

#### Table 1 – Building site plant schedule

Equipment | Power (kW) | Number | kW | kVA | V | A |

Tower Crane | 75 | 1 | 75 | 93.75 | 415 | 130.4 |

Sump Pump | 5 | 1 | 5 | 6.25 | 240 | 26 |

Lamps | 0.1 | 60 | 6 | 6 | 240 | 25 |

Flood Lamps | 0.4 | 12 | 4.8 | 4.8 | 240 | 20 |

Hand Tools | 0.4 | 20 | 8 | 10 | 110 | 90.9 |

Total // | 98.8 | 120.8 | n.a. | n.a. |

#### Table 2 – Building site energy use

Equipment | Power (kW) | Number | Time (h) | Energy (kWh) |

Tower Crane | 75 | 1 | 480 | 36000 |

Sump Pump | 5 | 1 | 480 | 2400 |

Tungsten lamps, security Lamps | 0.1 | 20 | 5824 | 11648 |

Lamps | 0.1 | 40 | 300 | 1200 |

Flood Lamps | 0.4 | 12 | 300 | 1440 |

Hand Tools | 0.4 | 20 | 480 | 3840 |

Total Power Used // | 56528 |

The total energy used by the systems is found from:

**kWh = number of appliances × kW per appliance × operation hours as shown in Table 2. **

If electricity costs **8 p per unit (kWh)** then the estimated cost for the 1-year contract will be:

**Reference:** Building Services Engineering // Electrical Installations – David V. Chadderton (Purchase paperback from Amazon)

Rohit subramanian

I have problem in one of the site where the connecting load is High HP pump -150 and 60 with 500kva transformer connected .

the MD is 185 connected obtained but the ideal time is so high that power factor is getting low checked out the parameters additional 1.5KVAR given for the grid than also the power factor is not getting maintained.

last month MD shooted up to 193.4 KVA but the running of the plant was minimal .

Help out in this regards

ALI

hay i am doing analysis on base of MDI penalty can anyone please tell me that how it is calculated. in bill i find readings like this

PRV CUMM MDI 98 PRS CUMM MDI 104 MDI 6

what does this mean….

magdy

we just start a construction of abig project at ajyad of mekkah saudi arabia we wanna make calculation of the total power needed for service using demand and load factor please tel me me how to compute the demand factor

Munaf

Thank you for the information

K.Ruju

Thanks for the Information

jesus g

thanks

Loum Peter

I’m so sorry i had forgotten the principle of the power triangle so sorry really the 93.75kva for the crane is correct.

Bimal

Nice good and very helpful

Thomas

How about the power consumption for testing& commissioning?

Loum Peter

Thanks for your information, but one think..for three phase , power W=1.732xV(line)xI(line)xPF no? if so that means VA =W/(1.732×0.8) and the KVA for the crane would be 54.13KVA….

jesus perez

No. Active power (W) is equal to VA x power factor

sandar

Thanks u for ur information

Anthony Uwadiae

Thanks for the information.

Bonginkosi

Thank you Edvard Csanyi for an informative and summarized example.

it assisted me in a project I am doing now.

I keep reading your articles, these are very helpful articles

the whole page is very useful, keep up the good work